How do you multiply $(m - 1) (m^{2} + 2 m + 6)$ $(m - 1) (m^2 + 2m + 6)$ ?

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Answer 1 · 2015-04-13T11:04:00

First you multiply the $m$ $m$ with all the others in the second group, and then the $- 1$ $-1$

$= m \cdot (m^{2} + 2 m + 6) + (- 1) \cdot (m^{2} + 2 m + 6)$ $=m*(m^2+2m+6)+(-1)*(m^2+2m+6)$

$= (m . m^{2} + m \cdot 2 m + m \cdot 6) + (- 1 m^{2} - 2 m - 6)$ $=(m.m^2+m*2m+m*6)+(-1m^2-2m-6)$

$= m^{3} + 2 m^{2} + 6 m - 1 m^{2} - 2 m - 6$ $=m^3+2m^2+6m-1m^2-2m-6$ . now add like powers:

$= m^{3} + (2 - 1) m^{2} + (6 - 2) m - 6$ $=m^3+(2-1)m^2+(6-2)m-6$

$= m^{3} + m^{2} + 4 m - 6$ $=m^3+m^2+4m-6$

How do you multiply (m - 1) (m^2 + 2m + 6)(m−1)(m2+2m+6)(m - 1) (m^2 + 2m + 6)?